Convergence of $\left(\ln\left(1+ \frac{1} {x}\right)\right) ^n $

Question

Consider the following sequence of functions defined by $$ f_n(x)= \left[\ln\left(1+ \frac{1} {x}\right)\right] ^n $$ with two different domains, first for $x\in (0,1]$ and then $x\in [\frac{1} {2} ,1]$ What can I say about its convergence?

I've found it similar to the geometric sequence, but I'm not very sure.
I find the problem a bit puzzling. I don't see a difference between the two intervals (in this type of problems the sequence behaves differently depending on the compactness/openness of the interval). Please check it and feel free to give any comment.

Answer 1

My attempt

$$ - 1 \lt \ln\left(1+ \frac{1} {x}\right) \le 1 $$ for $$ 1+ \frac{1} {x} \in \left( \frac{1} {e}, e \right] $$ which happens when $$ x\ge \frac{1} {e-1}, x\lt \frac{e} {1-e}$$

Restricting to the values in the domains of the problem, we have $\frac{1} {e-1} \gt \frac{1} {2}$, so I can say that the sequence converges pointwise but not uniformly for $ x \in [\frac{1} {e-1}, 1]$. For the other values $x \in (0, \frac{1} {e-1}) $ the sequence does not converge.

Answer 2

It is clear that the sequence $(f_n(x))_{n\in\Bbb{N}}$ converges if and only if $$-1<\ln\big(1+\tfrac1x\big)\leq1.$$ Of course given that $x>0$ we have $\ln(1+\tfrac1x)>0$, and $\ln(1+\tfrac1x)\leq1$ if and only if $x\geq\tfrac{1}{e-1}$.